Memcal installation direction

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Jul 15, 2013 | 05:49 PM

dirtyrobinson

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From: Central MN

Memcal installation direction

Does anyone know the correct orientation of the memcal chip for proper installation in a 1985 1226870 ECM? I don't wanna install my new chip backwards and I forgot to check the orientation of the old one before pulling it out.

Also, I'm curious if anyone knows what type of PROM chip these are? EPROM? Size in kb? Part no.? Can I find these from an electronics store or are they specially designed chips?

Last edited by dirtyrobinson; Jul 15, 2013 at 05:52 PM.

Jul 15, 2013 | 05:56 PM

dirtyrobinson

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Nevermind, I guess I'm not observant enough to realize that the housing is designed to only go in one way. This thread can get deleted if necessary. :-P

Jul 15, 2013 | 05:59 PM

MrWillys

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From: Reno Nevada

It is a 2732C chip and is 32k.

Jul 15, 2013 | 06:07 PM

dirtyrobinson

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Thank you very much for that information!

Jul 16, 2013 | 08:41 PM

vetteoz

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Quote:

Originally Posted by dirtyrobinson

This thread can get deleted if necessary.

FWIW
You can delete your own posts through the delete option in the EDIT button ;bottom right

Jul 17, 2013 | 11:24 AM

bjankuski

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From: Glenbeulah Wi

Quote:

Originally Posted by MrWillys

It is a 2732C chip and is 32k.

Minor correction, The chip is a 2732C but is is only 4K.

Jul 17, 2013 | 06:51 PM

MrWillys

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From: Reno Nevada

Quote:

Originally Posted by bjankuski

Minor correction, The chip is a 2732C but is is only 4K.

I've only known this as a 32k, and many others consider this also. Is it possible the .bin file is only 4k and where this discrepancy is?
http://www.datasheet4u.net/datasheet...uctor.pdf.html

Jul 17, 2013 | 09:50 PM

Cliff Harris

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From: Anaheim CA

It's 4K x 8 = 32K.

4K is the number of bytes. 32K is the number of bits (each byte is 8 bits).

Another way to put it: there are 32768 memory cells.

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Jul 17, 2013 | 11:31 PM

MrWillys

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From: Reno Nevada

The byte died with the FAT system which has been taken over by NTFS which has no minimum allocation?

Jul 18, 2013 | 08:57 AM

#10

bjankuski

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From: Glenbeulah Wi

Quote:

Originally Posted by Cliff Harris

It's 4K x 8 = 32K.

4K is the number of bytes. 32K is the number of bits (each byte is 8 bits).

Another way to put it: there are 32768 memory cells.

I agree with your statement, I use offsets when I program chips and I use 4096 bytes as the chip size. I do not use bits, so depending on what the question was I understand either answer.

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